Practice Test 2 - Chem 101

The compound SiO is called silicon monoxide

TRUE

Consider the adjacent molecule. I left out the lone pairs. Select a true statement a) The angle C2-C1-O should be about 109o. b) The oxidation number of C4 is +2 c) There are 9 lone pairs missing d) C3 is best described with sp3-hybridization e) The formal charge of N is +2

B OR C A- 120 DEGREES D- SP HYBRIDIZATION E- FALSE

Question 1 The image to the left symbolizes a) The formation of a bonding π molecular orbital b) The formation of an antibonding π* - molecular orbital c) The formation of an antibonding σ* molecular orbital d) The formation of a non-bonding lone pair e) The formation of a σ-bonding molecular orbital

A

Consider the ion BrO21- You may assume the bromine atom is in the middle. Perform a Lewis type calculation. Determine the formal charges (FC), oxidation states (Ox) and the bond order (BO). Also consider the shape and the proper name of the molecular ion. Select the correct answer: a) The shape is bent, the name is the bromite and Ox(Br) = +3 b) The shape is linear and Ox(Br) = -1 c) The formal charges are: FC(Br)=+3 and FC(O)=-2 d) There are ten lone pairs and the name is bromate. e) There are two resonance structures, such that the bond order of both Br-O bonds is 1½.

A B- SHAPE IS BENT C- FORMAL CHARGE ON B IS +1 D- 7 LONE PAIRS AND ITS BROMITE E - NOT TWO RESONANCE STRUCTURES

Select a true statement a) Resonance structures are Lewis structures that differ only in the placement of their electrons b) The process that takes a molecule from one resonance structure to the other is relatively slow c) In the two resonance structures of SO2: O-S=O ↔ O=S-O the formal charges remain the same but the oxidation numbers change for the oxygen atoms d) The percent ionic character of a bond is determined by the difference in atomic mass of the two atoms e) 5d orbitals can contain more electrons than 3d orbitals

A b- completely false; its rapidly moving back and forth c - false d - ionic character is determined by how ionic a bond is e- false; all d orbitals are equal

Which of the following contains the largest number of π-electrons? a) 673 g of CH4 gas b) 1 mole of nitrogen gas N2 c) 1 mole of ethene gas H2C=CH2 d) 18 grams of water vapor e) All of the above contain about the same number

B

In the above diagram an oxygen molecule is represented in term of its molecular orbitals on the left. It is exposed to a powerful light beam and a photochemical process occurs. One of the π* electrons makes a transition to become a free electron, resulting in the species on the right. A. Both species are paramagnetic, the total spin on the left is S=1, on right S= ½ , resulting in a triplet on the left and a doublet on the right, respectively B. The two pairs shown in the lowest two molecular orbitals represent lone pairs, localized one on each atom. C. The species on the right has one antibonding electron less, its bond order increases from 2 to 2½ resulting in a shorter bond D. The excitation destroys the bonding so that the species on the right would quickly fall apart into an oxygen atom and an oxygen anion E. The number of bonding electrons on the left is larger than on the right

A AND C B - THE LOWEST MOLECULAR ORBITALS DONT REPRESENT LONE PAIRS AND THEY AREN'T LOCALIZED D- EXCITATION DOESN'T DESTROY THE BONDING

Molecules are not always easy to draw on a 2D white paper like the one you are looking at. The ammonia molecule can be represented in various ways, four of which (P.Q.R.S) are shown to the left. Select a true statement a) R and S are representations that do not depend on any theories of bonding like Lewis or VSEPR, but reflect the shape as determined experimentally b) In P all the N-H bonds are shown as equally long; this is incorrect and in Q the difference in length is shown more correctly c) R is called "space-filling" whereas S is called "ball-and-stick" d) The "line-wedge-dash" representation reflects VSEPR theory and the 3D nature of the molecule that Lewis theory originally ignored e) The space-filling S representation exaggerates the size of the atoms

A AND D B- FALSE; ALL BONDS ARE SAME LENGTH IN P AND IN Q DIFFERENCE IN SPACE NOT LENGTH IS SHOWN C- FALSE; R IS BALL AND STICK AND S IS SPACE FILLING E- FALSE; SPACE FILLING DOESN'T EXAGGERATE SIZE

John studies the reaction: MnCl4 + 2 FeCl2 --> MnCl2 + 2 FeCl3 1 : 2 :: 1 : 2 He mixes 19.67 g MnCl4 and 25.35 g FeCl2 and heats up this mixture under an atmosphere of pure argon. What do you expect him to get? a) 32.44 g FeCl3 and one other compound b) 16.22 g FeCl3 some MnCl2 and some unreacted MnCl4 c) 12.58 g MnCl2 some FeCl3 and some unreacted FeCl2 d) 12.58 g MnCl2 and one other compound e) None of the above

A OR D First express quantities in moles: 19.67[g]/196.74[g/mol]= 0.1 mole MnCl4. 25.35[G]/126.75[g/mol]= 0.2 mole. Dividing the coefficients all by 10 we get 0.1 : 0.2 :: 0.1 : 0.2 So, we do not have a limiting reactant: reagents are present in the right ratio. The reaction will leave no reagents and we will get 0.1 [mol] MnCl2 which converts into 0.1*125.84[g/mol]= [[[[[[[[12.58 [g] MnCl2]]]]]]]]] 0.2 [mol] FeCl3 which converts into 0.2*162.2 [g/mol] = [[[[[[[[[[[32.44 [g] FeCl3]]]]]]]]]]

What are the correct names for these species?

B

Consider the molecular orbital diagram of the hypothetical HeH2 molecule shown on the left and select a true statement a) The electron pair denoted Q represents a sigma bond between the two hydrogen atoms (Non-bonding pair) b) The bond order or each He-H bond is only ½ c) The R-level represents an empty π*- orbital d) Because only 1s orbitals are involved in the formation of the MO's the only nodal planes are the confinement nodes at infinity e) The bonding orbital P is mostly localized on the helium atom because its Zeff-value is higher

B OR E a) The electron pair denoted Q represents [/a sigma bond/] between the two hydrogen atoms (ACTUALLY A Non-bonding pair) c)FALSE The R-level cannot represent an empty π*- orbital because HE DOESN'T HAVE PI BONDS WITH HYDROGEN d) Because only 1s orbitals are involved in the formation of the [/MO's the only nodal planes/] are the confinement nodes at infinity

Consider the hypothetical molecule BCN from a perspective of Lewis theory, including formal charges and select a true statement The Lewis calculation gives SP=6 and LP=0, but boron is electron deficient so maybe that doesn't give a good picture. a) The molecule is best described by the Lewis structure B≡C≡N b) The most reasonable electronic structure would be |B-C ≡N| c) The most reasonable electronic structure is |B=C=N| Here we do have non-zero FC's d) The B-C-N bond angle is 120o, because of the lone pair on the carbon atom e) None of the above

B; The Lewis calculation gives SP=6 and LP=0, but boron is electron deficient so maybe that doesn't give a good picture. a) The molecule is [/best described by the Lewis structure B≡C≡N/] With 6 bonds on carbon?! b) The most reasonable electronic structure would be |B-C ≡N| No octet. but all FC=0 c) The most reasonable electronic structure is |B=C=N| Here we do have non-zero FC's d) The B-C-N bond angle is 120o, [/because of the lone pair on the carbon atom/]

Consider the shape of the following molecules and select the true statement

C

Consider the Lewis structures of the acetic acid molecule (CH3COOH) and its acetate (CH3COO-) ion. Select a true statement a) In the acid there are two different C-O distances, whereas in the acetate ion the bond lengths are the same b) The acetate ion possesses two resonance structures, the acetic acid molecule only one c) In both molecules, the atoms C(left) and C(right) are best described with sp3 (left) and sp2 (right) hybridization d) All of the above e) None of the above

D

Select a true statement a) Hybridization theory works well in describing the bonding in most organic molecules. b) Nonbonding electron pairs are often thought to be associated to more than atom, whereas lone pairs are assumed to be localized on one atom only. c) The oxygen molecule's paramagnetic properties can be explained by molecular orbital theory, but not by Lewis, VSEPR or hybridization theory d) All of the above. e) None of the above.

D

Select a true statement a) Lewis theory provides a good understanding of the bonding in diatomics like N2, O2, CO and NO b) Overlap between orbitals is important in Valence Bond theory only, not in Molecular Orbital theory c) Triple and double bonds are an important aspect of Ionic Bonding d) Molecular Orbital theory allows for bonding between more than two atoms simultaneously. Lewis theory does not. e) The p-orbitals of an atom can only contribute to π-type overlap

D

a) All but four of the carbon atoms in the above molecule should be described with sp2 hybridization b) The bond length of A is about the same as the bond length of C c) The bond length of B is shorter than the bond length of D d) Both a) and c) are true e) All of the above is true

D

When calculating the electronic structure of the carbonate ion CO32- according to the Lewis assumptions we arrive at three equivalent structures as shown. The (formal) charges are not shown. Select the correct statement a) The description with three resonance forms results from the erroneous assumption that electron pairs (e.p.) can only be shared by two atoms. In reality the bond order is 1.333 for each bonding region, because one e.p. is shared over the whole molecule. b) The oxidation state for the carbon atom is -4 in each of the resonance structures c) In each of the structures the formal charge on the singly bound oxygen atoms is -1 d) Both a and c are true. e) Both b and c are true

D B IS FALSE- ITS +4

Carbon suboxide C3O2 is a molecular compound. The molecule has the form of a chain. Do a Lewis type calculation and select a true statement a) There should be 5 lone pairs to satisfy the Lewis postulates b) The most stable electronic structure is the resonance pair O-C ≡C-C ≡O|↔| O≡C -C≡C- O c) The oxidation number for each of the carbon atoms is 1.333 d) All C-atoms are best described by sp-hybridization; all O-atoms by sp2-hybridization e) All of the above ER: 5*8 =40 VE: 3*4+2*6=24 12 TP ----------------- - 16 8 SP ---------------- 4 LP Best: <O=C=C=C=O> all FC zero

D ER: 5*8 =40 VE: 3*4+2*6=24 --> 12 TP ----------------- - 16 --> 8 SP ---------------- 4 LP Best: <O=C=C=C=O> all FC zero

Consider the azide ion N3- . Do a Lewis calculation and determine if there are any resonance structures. Use formal charges to determine which electronic structure(s) would be the most stable one(s). Select a true statement. a) The ion has a bent shape. The N-N-N angle is about 120o. b) The most stable structures are the resonance pair N-N≡N ↔ N≡N-N c) The species has an odd number of electrons, which makes a Lewis calculation difficult d) The most stable electronic structure is the N=N=N one, even though its formal charges are not minimal e) The formal charges are equal to the oxidation numbers

D AND E A- AZIDE ION IS LINEAR B- NO RESONANCE C- FALSE; THE SPECIES HAS ODD NUMBER BUT THE ION ADDS ANOTHER ELECTRON

The diagram to the left depicts the molecular orbitals formed by the pz-orbitals of a hypothetical linear B4 molecule. The bonding caused by the boron atom's s-electrons can be neglected. The MO's are not shown as ordered by their increasing energy. Select a true statement. a) Only one MO, the one marked "b" would be filled b) Both "a" and "b" would be filled, leading to a bond order of 2/3 c) The energy of "c" is lower than that of "d" d) The "b" orbital has two nodal planes e) None of the above is true

D; Boron has 2s22p1 so there is only one p-electron to play with. Only the a) orbital will be filled I gave partial for e) a) and b). The right answer is d) the "b"-orbital has one 'extra node' between the two middle atoms and 'inherits' a nodal plane from the fact that is made up of p-functions

The molecule PCl5 has an interesting shape as shown to the left. Select a true statement a) This shape is called a trigonal bipyramid b) The five chlorine atoms (the ligands) are not all equivalent c) There are two different Cl-P-Cl bond angles: 90 and 120o d) This molecule defies the 'octet rule' e) All of the above

E

What are the proper names for these compounds:

E

Which of the following contains the highest number of nucleons? a) 12 g of carbon b) 12 g of hydrogen c) 12 g of acetic acid CH3COOH d) 12 g of horse meat e) All of the above contain about the same number

E

The bonding in carbon monoxide is better described by the Bohr model than by Lewis theory

FALSE

Triple bonds consist of three bonds of equal strength

FALSE (No ;one stronger sigma and two weaker pi-bonds)

Molecular orbitals combine atomic orbitals on the same atom, whereas hybridization combines atomic orbitals on adjacent atoms

FALSE Molecular orbitals combine atomic orbitals on [/the same atom/], whereas hybridization combines atomic orbitals on [/adjacent atoms/]

Heterodiatomic molecules have their HOMO above the LUMO in energy

FALSE Heterodiatomic molecules are molecules with two elements that are different. eg HF

In π-bonding most of the electron density can be found on the axis that connects the two nuclei

FALSE; AXIS THAT CONNECTS NUCLEI IS SIGMA BONDING

Ionic bonding cannot be described using molecular orbitals

FALSE; IT CAN

A triatomic molecule will only have a bent shape if its middle atom carries one lone pair

FALSE; IT CAN BE BENT WITH TWO LONE PAIRS ON THE CENTRAL

The energy of a pair of atoms generally depends on their distance R and has a minimum for R is approximately 1 - 3 nm.

FALSE; NO MINIMUM

Nodal planes separate regions of opposite charge

FALSE; NO!!! Opposite phase.

1 μmol of mercury contains about 6.022 10+20 atoms (Off by a factor of 1000)

FALSE; OFF BY A FACTOR OF 1000

Covalent bonding is directional because shared pairs repel each other. In ionic bonding there are no shared pairs

TRUE

In a heterodiatomic molecule the difference in electronegativity often causes the presence of a dipole

TRUE

Neither lone pairs nor non-bonding pairs contribute to bonding, but non-bonding electrons are not limited to one atom only

TRUE

π-bonding cannot occur between two s-orbitals, but it can between a d-orbital and a p-orbital

TRUE

Singlet oxygen is an excited state of triplet oxygen

TRUE; The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.